| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2004 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Find equation satisfied by limit |
| Difficulty | Standard +0.3 This is a straightforward fixed point iteration question requiring routine application of the formula (part i) and simple algebraic manipulation to find the equation satisfied by the limit (part ii). The key insight that x_{n+1} = x_n at convergence is standard, and rearranging to show α^5 = 306 involves only basic algebra. Slightly easier than average due to its mechanical nature and clear structure. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use the given iterative formula correctly at least once with \(x_1 = 3\) | M1 | |
| Obtain final answer 3.142 | A1 | |
| Show sufficient iterations to justify its accuracy to 3 d.p. | A1 | Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State any suitable equation e.g. \(x = \frac{1}{5}\left(4x + \frac{306}{x^4}\right)\) | B1 | |
| Derive the given answer \(\alpha\) (or x) \(= \sqrt[5]{306}\) | B1 | Total: 2 |
## Question 2:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use the given iterative formula correctly at least once with $x_1 = 3$ | M1 | |
| Obtain final answer 3.142 | A1 | |
| Show sufficient iterations to justify its accuracy to 3 d.p. | A1 | **Total: 3** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| State any suitable equation e.g. $x = \frac{1}{5}\left(4x + \frac{306}{x^4}\right)$ | B1 | |
| Derive the given answer $\alpha$ (or x) $= \sqrt[5]{306}$ | B1 | **Total: 2** |
---2 The sequence of values given by the iterative formula
$$x _ { n + 1 } = \frac { 1 } { 5 } \left( 4 x _ { n } + \frac { 306 } { x _ { n } ^ { 4 } } \right)$$
with initial value $x _ { 1 } = 3$, converges to $\alpha$.\\
(i) Use this iterative formula to find $\alpha$ correct to 3 decimal places, showing the result of each iteration.\\
(ii) State an equation satisfied by $\alpha$, and hence show that the exact value of $\alpha$ is $\sqrt [ 5 ] { 306 }$.
\hfill \mbox{\textit{CAIE P2 2004 Q2 [5]}}